To win at BIG GAME in a certain state, one must correctly select 6 numbers from a collection of 49 numbers (1 through 49). The order in which the selections is made does not matter. How many different 6-number selections are possible, assuming no repetition of numbers?
Choose one answer.
number of choices,no repetition?
49*48*47*46*45*44 or more likely your teacher wanted you to use the formula
choices: 49!/(49-6)!
if 11,47,81,18,65, were selected from 1-90 ,what are the next numbers to be selected
if 11,47,81,18,65, were selected from 1-90 ,what are the next five numbers to be selected
To determine the number of different 6-number selections possible, we can use the concept of combinations. In this case, we need to calculate the number of combinations of 49 numbers taken 6 at a time.
The formula for combinations is given by:
C(n, r) = n! / (r!(n - r)!)
Where:
C(n, r) represents the number of combinations of n items taken r at a time,
n! represents the factorial of n (the product of all positive integers from 1 to n),
r! represents the factorial of r,
and (n - r)! represents the factorial of (n - r).
In this case, we have:
n = 49 (the total number of numbers)
r = 6 (the number of numbers to be selected)
Plugging in the values into the formula, we get:
C(49, 6) = 49! / (6!(49 - 6)!)
Simplifying the equation further, we have:
C(49, 6) = 49! / (6!43!)
Calculating the factorials, we get:
C(49, 6) = 49! / (720 * 600766864000)
Using a calculator, we evaluate 49! to be approximately 6.16 x 10^57.
Therefore, the number of different 6-number selections possible without repetition is:
C(49, 6) = 6.16 x 10^57 / (720 * 600766864000)
Calculating this value, we get:
C(49, 6) ≈ 13,983,816
Hence, there are approximately 13,983,816 different 6-number selections possible in the given scenario.